[Ari Ben Canaan]

What volume of 0.2M NaOH is required to prepare 200ml of a 0.2M buffer solution of pH 4 (pKa of acid is 3.5)?

So the buffer equation is :

HA ------> H+ + HA-

I started off by calculating the concentration of H+ ions that would cause a pH of 4.

I got 1*10-4

Then I calculated the Ka (3.162*10-4)

3.162x10-4 = [H+]² / [HA]
Hence, 3.162x10-4 * 1x10-4 = [HA]

[HA] = 3.16x10-5

At this point I have no idea what to do.

[Borek]

So after adding the NaOH the equation is :

HA ------> H+ + HA-

No, dissociation occurs even when there is no base present.

Acid does react with base - can you write the reaction equation?

This is a similar question (not identical), see if it will help:

http://www.chembuddy.com/?left=pH-calculation-questions&right=pH-buffer-q1

[Ari Ben Canaan]

Yikes !! I meant to say that was my general buffer equation.... I've edited the OP

EDIT : Ignore this post

[Ari Ben Canaan]

What do they mean by 'a 0.2M buffer solution of pH 4' ?

What does the 0.2M concentration refer to ?

[Borek]

Sum of concentrations of both acid and conjugate base equals 0.2M.

[Ari Ben Canaan]

Sum of concentrations of both acid and conjugate base equals 0.2M.

Okay I just want you to check if I'm headed in the correct direction. I'm just taking the acid to be ethanoic acid as I prefer dealing with an actual formula.... I know that the acid may not be ethanoic acid.

NaOH + CH3COOH  CH3COOH- Na+  + H2O

CH₃COOH  CH₃COO^- + H⁺

[CH₃COOH] + [CH₃COO^-] = 0.2

So if [CH₃COO^-] = x  [CH₃COOH]  = 0.2-x

Using Ka :

10^-3.5 = 10^-4*x  /  0.2-x

x = 0.1519

Where do I go from here ?

[Borek]

x = 0.1519

Now it is just a matter of converting this value to volume.

[Ari Ben Canaan]

x = 0.1519

Now it is just a matter of converting this value to volume.

Ah !!! I see.... 0.1519 * (200/1000) = 0.03038 moles in 200 ml.

(0.03038*1000) divided by 0.2 = 151.9 ml ?

[Borek]

Wasn't that hard.